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Since the space $E:=\mathcal M_n(\mathbb C)$ of all $n \times n$ complex matrices is a finite-dimensional vector space, all norms define the same topology. So we can take a sub-multiplicative norm, that is, a norm $\lVert\cdot\rVert$ such that $\lVert AB\rVert\leq \lVert A\rVert \cdot\lVert B\rVert$. (For example, we can take $\lVert\cdot\rVert$ to be the operator norm on $E$.) As a finite dimensional vector space, $E$ is complete, so it's enough to show normal convergence. We have that, for each integer $n \geq 0$,
$$0\leq \lVert\frac{A^n}{n!}\rVert\leq \frac{\lVert A\rVert^n}{n!},$$
and we know that, for each real number $x$, the series $\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ converges (it defines the exponential function). Therefore, for any $A \in E$, the series $\sum_{n=0}^{+\infty}\frac{A^n}{n!}$ converges. (We also got the additional result that $\lVert e^A\rVert\leq e^{\lVert A\rVert}$ for any $A \in E$.)